How to solve tanx + cotx = 4cos2x? Thank you

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Answer 1 · 2018-02-18T11:01:12

General solution: $x=(4n+1)pi/8 ; n in z$

$tanx+cotx=4cos2x or sinx/cosx+cosx/sinx=4cos2x$ or

$(sin^2x+cos^2x)/(cosx*sinx)=4cos2x$ or

$1/(1/2(2cosx*sinx))=4cos2x ; [sin2x=2sinxcosx]$ or

$2/(sin2x)=4cos2x or 2sin2xcos2x=1$

$[sin4x=2sin2xcos2x]$ or

$sin4x=1; sin (pi/2)=1 :. 4x =pi/2 :. x=pi/8$

General solution: $x=(4n+1)pi/8 ; n in z$ [Ans]